Q. A metal rod of resistance 20 Ω is fixed along a diameter of a conducting ring radius 0.1 m and lies on x-y plane. There is a magnetic field $ \displaystyle \vec{B} = 50 \hat{k} T $. The ring rotates with an angular velocity ω = 20 rod s^{-1} about its axis. An external resistance of 10 Ω is connected across the center of the ring and rim. The current through external resistance is

(a) 1/4

(b) 1/2

(c) 1/3

(d) 0

Ans:(c)

Sol:

Potential difference between centre of the ring and the rim is

$ \displaystyle E = \frac{1}{2}B \omega r^2 $

$ \displaystyle E = \frac{1}{2}\times 50 \times 20 \times (0.1)^2 $

E = 5 volt

$ \displaystyle r = \frac{10 \times 10}{10 +10} $

r = 5 Ω

Current through external resistance

$ \displaystyle i = \frac{E}{R + r } = \frac{5}{10 + 5} $

i = 1/3 A